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Calorimetry: Latent Heat of Vaporization©98 Experiment 6 Objective: To determine the heat of vaporization of liquid nitrogen. DISCUSSION: A calorimeter is a device that allows accurate determination of energy and temperature changes during simple thermodynamic processes. A key requirement is that it can act as an isolated, closed system. Many very sophisticated types of calorimeters exist. One very common type is known to scientists as the Dewar flask, or simply ‘Dewar’. You know it as the Thermos bottle. An even simpler calorimeter is a Styrofoam cup. It is a satisfactory isolated system for use in small experiments, such as the one that is to be conducted here. The science of making these measurements is referred to as calorimetry. Experiments in calorimetry are quite common in high school physics and chemistry labs. The prevalent experiments are these two: the specific heat, c, of an unknown metal; and the latent heat of fusion (solidification), LF, of water into ice. Instead of duplicating those standard experiments, we will attempt to try and find a more unusual constant: the latent heat of the vaporization of nitrogen, Lv(nitrogen). Let us review the standard process of heating ice from -20° Celsius to superheated steam at 120° Celsius: a. Heat the ice at - 20 to 0° Celsius. This heating process involves no change of phase, so the heat required is expressed by Q = mciΔT, where m is the mass of the ice, ci is the specific heat of ice, and T is the temperature change of the sample. The specific heat of ice is interestingly close to half of that of water. Ice has a specific heat ci= 2,050 J/kg-K, while water has a specific heat of cw= 4,190 J/kgK. Recall that changes in temperature are the same on the Celsius and Kelvin scales: ΔT is 20°Celsius, which is the same as 20 Kelvin. So given a mass of ice, m, the heat required to warm the ice is Q(1) is 41,000m. b. Step 2 is the phase change from ice into water. The heat is expressed by Q(2) =mL. We can look up L and find that it is 334,000 J/kg. So Q(2) is 334000m. c. Heat water from 0 to 100° C. So, using Q = mcΔT, we have Q(3) = 419,000m. d. Another phase change. The water at 100°C is turned into steam at 100°C, and given the latent heat of the vaporization of water, Lv is easy to calculate, Q(4) =mLv =2,257,000m. e. Heating the steam from 100°C to 120°C. Using steam’s c in this range, 2,000J/kg-K, we have Q(5) = mcΔT =40,000m. 6-1 The total heat required for the process is given by: QTotal Q(1) Q( 2) Q(3) Q( 4) Q(5) QTotal m41,000 334,000 419,000 2,257,000 40,000 So it will be easy to determine any thermodynamic process using the proper analysis of the materials involved regarding phase, specific heat, temperature, etc. Let us now look at the process we can use to find a fairly unusual constant, the latent heat of vaporization of liquid nitrogen. We will exploit the well known thermodynamic properties of water. We can use hot water to vaporize liquid nitrogen. Thanks to the conservation of energy, we know that the heat given off by the water is exactly the heat absorbed by the nitrogen. In other words, the total Q of the system is constant. Instead of writing an expression of the type Q lost by materials + Q gained by materials = 0, it is more traditional to state: heat lost by hotter materials = heat gained by cooler materials . For our process we have: heat lost by hot water = heat used to vaporize nitrogen Using the temperature change relation Q = mcΔT and Q = mL in the appropriate places we have: mwater cwater T water mnitrogen Lv ( nitrogen) (1) where ΔTwater = | Ti-Tf | represents the temperature change of the water. EXERCISES: 1. Obtain an empty Styrofoam cup, and find its mass, mcup. 2. Take the cup to the teaching assistant who will add a small amount of liquid nitrogen (LN2). Return to your station and place the Styrofoam cup onto the balance to find the mass of the liquid nitrogen and cup. 3. Subtract the mass of the cup to determine the mass of the remaining liquid nitrogen, mnitrogen. 4. Take the second Styrofoam cup and obtain half a cup of hot water from the hot pot. Now record it’s mass and temperature, Ti. 6-2 5. Immediately but slowly pour the hot water into the liquid nitrogen. The nitrogen will boil rapidly. Be sure to hold the water above the escaping vapor while pouring (explainWHY?). 6. Once the vapor has subsided, find the mass and final temperature (Tf ) of the water, verifying that the nitrogen has completely evaporated. 7. Solve equation (1) for Lv(nitrogen). 8. Repeat the experiment two more times. 9. Find the percent difference between the standard value and each result as well as a comparison with the average of the results. What is the benefit of performing the experiment more than once? 6-3 Data Sheet – Experiment 6 Mass of cup mcup1 Mass of nitrogen and cup m(LN2+cup1) (kg) (kg) Mass of nitrogen Mass of cup and water Mass of water mnitrogen= m(cup2+water) mwater m(LN2+cup1) mcup1 (kg) (kg) (kg) Initial temp of water Final temp of water Ti Tf (ºC) (ºC) Latent heat of vaporization for nitrogen Lv(nitrogen) (J/kg) 6-4